# Investigate Each Limit....(A)

• nycmathguy
I am a math major.In summary, the conversation focused on investigating the limits of a piecewise function with the definition f(x)=1 if x is an integer and f(x)=0 if x is not an integer. The experts discussed how the function behaves at different intervals and how to determine the limit as x approaches different values, including integers and rational numbers. They also discussed how the function can be tweaked by giving a specific value to f(0) and how this affects the limit. The conclusion was that regardless of the value given to f(0), the limit of the function at x=0 will always be 0 as long as f(x)=0 for all x except 0.

#### nycmathguy

Homework Statement
Investigate each limit.
Relevant Equations
See attachment for function.
Investigate each limit.

See attachment.

1. lim f(x) x→2

2. lim f(x) x→1/2

I don't understand this piecewise function.

#### Attachments

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nycmathguy said:
Homework Statement:: Investigate each limit.
Relevant Equations:: See attachment for function.

Investigate each limit.

See attachment.

1. lim f(x) x→2

2. lim f(x) x→1/2

I don't understand this piecewise function.
Plot some points. What are f(0), f(1/4), f(1/2), f(2), f(2.5), etc.?

Also, you've posted a few threads just now with little or no work shown. That's a violation of forum rules. You have to show some effort. You have the formula for the function -- sketch a graph of it.

Focus at an interval [n,n+1] where n is an integer. Answer the following questions to help you understand how this function goes
1) What is f(n)
2) What is f(n+1)
3) What is f(x) for every ##x\in(n,n+1)## for example for x=(2n+1)/2 the midpoint of n and n+1.

Delta2 said:
Focus at an interval [n,n+1] where n is an integer. Answer the following questions to help you understand how this function goes
1) What is f(n)
2) What is f(n+1)
3) What is f(x) for every ##x\in(n,n+1)## for example for x=(2n+1)/2 the midpoint of n and n+1.
Sorry but I don't get it. Still lost.

what is f(1) and f(2) equal to for example? Hint: 1 and 2 are integers

nycmathguy
The definition of the function f(x) tells you that f(x)=1 if x is integer and f(x)=0 if x is not integer.

nycmathguy
Delta2 said:
The definition of the function f(x) tells you that f(x)=1 if x is integer and f(x)=0 if x is not integer.

I say for (1), the answer is 0.
The answer for (2) is 1.

Yes?

nycmathguy said:
I say for (1), the answer is 0.
The answer for (2) is 1.

Yes
Νο, ##f(n)=f(n+1)=1## for all integers n. The function definition tells us that f(x)=1 if x is integer.

Delta2 said:
Νο, ##f(n)=f(n+1)=1## for all integers n. The function definition tells us that f(x)=1 if x is integer.
For (1), x tends to an integer. Thus, then f(x) = 1.

For (2), x tends to a rational number. Thus, f(x) = 0.

Yes but as x tends to an integer, it passes from all sorts of rationals and irrationals (from the left and right of integer) for which f(x)=0.

Delta2 said:
Νο, ##f(n)=f(n+1)=1## for all integers n. The function definition tells us that f(x)=1 if x is integer.
What about the following two cases using the same attachment?

Investigate each limit.

1. lim f(x) x→3

2. lim f(x) x→0

For (1), x tends to an integer. Thus, f(x) = 1.

For (2), x tends to 0, which is not an integer.
Thus, f(x) = 1.

Yes?

If I give you the following definition for f:
f(1)=f(0)=1
f(x)=0 for all x inbetween 0 and 1.

Then what do you think is the ##\lim_{x\to 0} f(x)## (or ##\lim_{x\to 1} f(x)##..

nycmathguy
nycmathguy said:
What about the following two cases using the same attachment?

Investigate each limit.

1. lim f(x) x→3

2. lim f(x) x→0

For (1), x tends to an integer. Thus, f(x) = 1.

For (2), x tends to 0, which is not an integer.
Thus, f(x) = 1.

Yes?
For both cases the limit is 0. (0 is an integer btw).

nycmathguy said:
Homework Statement:: Investigate each limit.
Relevant Equations:: See attachment for function.

Investigate each limit.

See attachment.

1. lim f(x) x→2

2. lim f(x) x→1/2

I don't understand this piecewise function.

Delta2 said:
If I give you the following definition for f:
f(1)=f(0)=1
f(x)=0 for all x inbetween 0 and 1.

Then what do you think is the ##\lim_{x\to 0} f(x)## (or ##\lim_{x\to 1} f(x)##..
Can you elaborate a little more?
It's just not sinking in. In fact, Sullivan stated in his book that this is considered a challenging problem.

hm ok let me see
If I tell you that f(x)=0 for all x then what is the ##\lim_{x\to 0} f(x)##.

nycmathguy
I think you are confusing the ##\lim_{x\to x_0}f(x)## with the ##f(x_0)##. These two are equal only if the function f is continuous at ##x_0##. But in this problem here we have to deal with a function f that is not continuous at every integer.

nycmathguy
Delta2 said:
hm ok let me see
If I tell you that f(x)=0 for all x then what is the ##\lim_{x\to 0} f(x)##.
In that case, it is 0.

Delta2
nycmathguy said:
In that case, it is 0.
Correct now let's say I tweak the function and the function f is now f(x)=0 for all x EXCEPT for x=0 which I define to be f(0)=1. Do you think that the above limit changes or remains the same?

Delta2 said:
Correct now let's say I tweak the function and the function f is now f(x)=0 for all x EXCEPT for x=0 which I define to be f(0)=1. Do you think that the above limit changes or remains the same?
You said except for x = 0. I say the limit is 1?

nycmathguy said:
You said except for x = 0. I say the limit is 1?
Nope it isn't 1. What is f(x) equal to ,as x tends to 0, for example what is f(0.5), f(0.4), f(0.3) , f(0.2) and so on..

nycmathguy
Delta2 said:
Nope it isn't 1. What is f(x) equal to ,as x tends to 0, for example what is f(0.5), f(0.4), f(0.3) , f(0.2) and so on..
So, f(every decimal number you listed) = 0 because decimal numbers are rational and rational numbers are not integers.

Delta2
nycmathguy said:
So, f(every decimal number you listed) = 0 because decimal numbers are rational and rational numbers are not integers.
That's correct. So what conclusion can you make from this? where does f(x) tend to as x tends to 0?

Delta2 said:
That's correct. So what conclusion can you make from this? where does f(x) tend to as x tends to 0?
So, f(x) tends to 0 as x-->0.

Delta2
nycmathguy said:
So, f(x) tends to 0 as x-->0.
yes and this is true regardless of what value we choose to give to f(0). As long as f(x)=0 for all ##x\neq 0## .

nycmathguy
Delta2 said:
yes and this is true regardless of what value we choose to give to f(0). As long as f(x)=0 for all ##x\neq 0## .
Trust me, I plan to journey through calculus l,ll, and lll. We will see limit questions up the wall.

Just to check your understanding, if i tell you f(x)=5 for all ##x\neq 0## and f(0)=10, what is the limit of f(x) as x tends to 0?

Delta2 said:
Just to check your understanding, if i tell you f(x)=5 for all ##x\neq 0## and f(0)=10, what is the limit of f(x) as x tends to 0?
This one is tricky.
I say the limit is 5.

Delta2
nycmathguy said:
For (1), x tends to an integer. Thus, then f(x) = 1.
No. f(x) = 1 if x is an integer, but for all other numbers, f(x) = 0.
The question is asking about ##\lim_{x \to 2} f(x)##, not f(x). Even though f(2) = 1, ##\lim_{x \to 2} f(x)## is some other value.
nycmathguy said:
For (1), x tends to an integer. Thus, f(x) = 1.
Again, no.
nycmathguy said:
For (2), x tends to 0, which is not an integer.
Thus, f(x) = 1.
First off, 0 is an integer. Second, you're again not distinguishing between function values (e.g. f(0)) and values of the limit. Here the limit expression is ##\lim_{x \to 1/2} f(x)##, which just happens to be the same as f(1/2).
nycmathguy said:
So, f(every decimal number you listed) = 0 because decimal numbers are rational and rational numbers are not integers.
Most "decimal" numbers are not rational (e.g., ##\pi \approx 3.141592## and ##\sqrt 2 \approx 1.414##), and some rational numbers are integers (e.g., 2/1, 6/2, and so on).

Delta2 said:
Just to check your understanding, if i tell you f(x)=5 for all x≠0 and f(0)=10, what is the limit of f(x) as x tends to 0?

nycmathguy said:
This one is tricky.
I say the limit is 5.
Right, but it's not tricky if you understand the idea of what a limit means.

Delta2
Mark44 said:
No. f(x) = 1 if x is an integer, but for all other numbers, f(x) = 0.
The question is asking about ##\lim_{x \to 2} f(x)##, not f(x). Even though f(2) = 1, ##\lim_{x \to 2} f(x)## is some other value.

Again, no.

First off, 0 is an integer. Second, you're again not distinguishing between function values (e.g. f(0)) and values of the limit. Here the limit expression is ##\lim_{x \to 1/2} f(x)##, which just happens to be the same as f(1/2).

Most "decimal" numbers are not rational (e.g., ##\pi \approx 3.141592## and ##\sqrt 2 \approx 1.414##), and some rational numbers are integers (e.g., 2/1, 6/2, and so on).
Right, but it's not tricky if you understand the idea of what a limit means.
Ok. There are many more limits coming our way in time. This is just the beginning of the long journey.

Delta2
nycmathguy said:
Ok. There are many more limits coming our way in time.
So make sure you understand the difference between, say, ##f(c)## and ##\lim_{x \to c} f(x)##. For a continuous function f, they will be the same, but not necessarily so for discontinuous or piecewise-defined functions.

nycmathguy
Mark44 said:
So make sure you understand the difference between, say, ##f(c)## and ##\lim_{x \to c} f(x)##. For a continuous function f, they will be the same, but not necessarily so for discontinuous or piecewise-defined functions.
Will do.

Delta2